# Side Proof 6

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This page will handle one of the long cases in the Human proof that completely multiplicative sequences have discrepancy greater than 3, so that the page can be shorter and not have so many long sections. Specifically, this page will take care of the case where we assume: f(2)=f(7)=f(19)=f(67)=1, f(23)=f(67)=-1.

## Proof

Updating the table:

0 1 2 3 4 5 6 7 8 9

0|+ + - + - - + + +   0-9
- - - - + + + - + +   10-19
- - - - - + - - + +   20-29
+ - + + - - + + + +   30-39
- - - + - - - + - +   40-49
+ + - - - + + - + -   50-59
+ - - + + + + - - +   60-69
-|? + ? + - + - + +   70-79
- + - - -|+ + - -|+   80-89
- - - + + - - ? + -   90-99
+ ? + ? - + - ? - ?   100-109
+ - + + - + + - - -   110-119
+ + - + - - + ? + -   120-129
+ ? + + - + - ? + ?   130-139


Now, because of the cut, f(71)+f(73)=0. We therefore have two cases:

## Case 1: f(71)=1, f(73)=-1

We have two equations: 1) f[145,154] = -4+f(149)+f(151) >= -4

2) f[295,304] = 5-f(101)+f(149)+f(151) <= 4

(1)-(2)+9: f(101) >= 1

Therefore, f(101)=1.

f[201,206] = 5+f(103), so f(103)=-1. f[207,218] = -7+f(107)+f(109)+f(211), so f(107)=f(109)=f(211)=1. s(116)=3+f(97), so f(97)=-1. However, now f[319,328] = -7+f(163), which forces the discrepancy above 3. Therefore f(71)=-1.

## Case 2: f(71)=-1, f(73)=1

Now, f[259,266] = 6+f(131)+f(263), so f(131)=f(263)=-1. f[469,474] = -5-f(157), so f(157)=-1. f[333,344] = 6+f(167)+f(337), so f(167)=f(337)=-1.

It seems we can get no further with this assumption, so we have to make a new assumption.

## Case 2.1: f(73)=f(97)=1, f(71)=-1

We have two inequalities to consider:

1) f[679,688] = 6-f(137)-f(227)-f(229)+f(683) <= 4

2) f[225,232] = 4+f(227)+f(229) <= 4

(1)+(2)-10: -f(137)+f(683) <= -2, so f(137)=1, f(683)=-1.

We have four inequalities to consider:

1) f[139,168] = -6+f(139)+f(149)+f(151)+f(163) >= -4

2) f[295,310] = 5-f(101)-f(103)+f(149)+f(151)+f(307) <= 4

3) f[603,616] = -5-f(101)+f(151)+f(307)+f(607)+f(613) >= -4

4) f[411,418] = -4+f(103)-f(139) >= -4

(1)-(2)+(3)+(4)+19: 2f(103)+f(151)+f(163)+f(607)+f(613) >= 4

Therefore, f(103) = 1. f[201,206]=5+f(101), so f(101)=-1. f[295,304] = 6+f(149)+f(151), so f(149)=f(151)=-1. f[141,168] = -7+f(163), which forces the discrepancy above 3. This assumption fails at 688. Therefore, f(97)=-1.

## Case 2.2: f(73)=1, f(71)=f(97)=-1

f[285,302] = 7+f(149)+f(151)+f(293), so f(149)=f(151)=f(293)=1. However, now f[149,168] = -7+f(163), which forces the discrepancy above 3.

This completes Side proof 6. The furthest this assumption could get was 688.